目录
- 反复涂色 (color)
参考实现
- @ 2026-5-31 9:55:35
#include<bits/stdc++.h>
using namespace std;
const int N = 1000005;
int h, w, n;
char a[N];
int d[N], q[N];
int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> h >> w;
n = h * w;
for (int i = 0; i < n; i++) d[i] = -1;
string t;
for (int i = 0; i < h; i++) {
cin >> t;
for (int j = 0; j < w; j++) a[i * w + j] = t[j];
}
int l = 0, r = 0;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
int id = i * w + j;
if (a[id] == '.') continue;
bool ok = 0;
for (int k = 0; k < 8; k++) {
int x = i + dx[k], y = j + dy[k];
if (x < 0 || x >= h || y < 0 || y >= w) continue;
int v = x * w + y;
if (a[v] == '.') ok = 1;
}
if (ok) {
d[id] = 0;
q[r++] = id;
}
}
}
if (r == 0) {
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) cout << '.';
cout << '\n';
}
return 0;
}
while (l < r) {
int u = q[l++];
int x = u / w, y = u % w;
for (int k = 0; k < 8; k++) {
int nx = x + dx[k], ny = y + dy[k];
if (nx < 0 || nx >= h || ny < 0 || ny >= w) continue;
int v = nx * w + ny;
if (d[v] != -1) continue;
d[v] = d[u] + 1;
q[r++] = v;
}
}
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
int id = i * w + j;
if (d[id] % 2 == 0) cout << '#';
else cout << '.';
}
cout << '\n';
}
return 0;
}
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反复涂色 (color)
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